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if $(X,\mathfrak{M},\mu)$ be a measure space and $\{A_i\}_{i\in I}$ be an uncountable family of disjoint measurable subsets of $X$ . What can be the relationship between $\mu(\underset{i\in I}{\bigcup}A_i)$ and $\underset{i\in I}{\sum}\mu(A_i)$?

where

$\underset{i\in I}{\sum}\mu(A_i)=\underset{\underset{\text{F is finite}}{F\subseteq I}}{\sup}\Bigg(\underset{i\in F}{\sum}\mu(A_i)\Bigg)$

Is there a special relationship between them?

Amirhossein Haddadian
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1 Answers1

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Of course we have to assume that $\bigcup_{i \in I} A_i$ is actually measurable. Then we have $\mu(\bigcup_{i \in I} A_i) \ge \sum_{i \in I} \mu(A_i)$, but there isn't much else that we can say. Certainly the inequality can be strict, and by simple examples like letting $A_i$ be the singletons in $\mathbb{R}$ with Lebesgue measure, we see that it can even read $\infty \ge 0$.

To see that we do have $\ge$, let $F$ be any finite subset of $I$. By writing $\bigcup_{i \in I} A_i$ as the disjoint union of the measurable sets $\bigcup_{i \in F} A_i$ and $\bigcup_{i \in I \setminus F} A_i$, we have $$\mu\left(\bigcup_{i \in I} A_i\right) = \mu\left(\bigcup_{i \in F} A_i\right) + \mu\left(\bigcup_{i \in I \setminus F} A_i\right) \ge \mu\left(\bigcup_{i \in F} A_i\right) = \sum_{i \in F} \mu(A_i)$$ and now take the supremum over all finite $F \subset I$.

Nate Eldredge
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  • Thank you for your reply. what can we say about their relashionship if the family of sets are not disjoint? – Amirhossein Haddadian Mar 26 '23 at 19:18
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    If the sets are not disjoint, it could also be the other way $\mu(\bigcup_{i \in I} A_i) < \sum_{i \in I} \mu(A_i)$. Indeed, this could happen even for a union of two sets. – GEdgar Mar 26 '23 at 19:24
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    @Amirhoseinhadadian: If they are not disjoint then I think all you can say is the very trivial $\mu(\bigcup_{i \in I} A_i) \ge \sup_{i \in I} \mu(A_i)$. – Nate Eldredge Mar 26 '23 at 19:42